a) Equal to its diameter
b) Half of its diameter
c) Double of its diameter
d) None of these

Answer: a
Explanation:
$$\eqalign{ & V = \pi {r^2}h{\text{ and }} \cr & S = 2\pi rh + 2\pi {r^2} \cr & \,\,\,\,\,\,\, = 2\pi r\left( {h + r} \right) \cr & {\text{Where, }}h = \frac{V}{{\pi {r^2}}} \cr & S = 2\pi r\left( {\frac{V}{{\pi {r^2}}} + r} \right) \cr & S = \frac{{2V}}{r} + 2\pi {r^2} \cr & \frac{{dS}}{{dr}} = \frac{{ – 2V}}{{{r^2}}} + 4\pi r{\text{ and}} \cr & \frac{{{d^2}S}}{{d{r^2}}} = \left( {\frac{{4V}}{{{r^3}}} + 4\pi } \right){\text{ > 0}} \cr} $$
S is minimum when :
$$\eqalign{ & \frac{{dS}}{{dr}} = 0 \cr & \frac{{ – 2V}}{{{r^2}}} + 4\pi r = 0 \cr & V = 2\pi {r^3} \cr & \pi {r^2}h = 2\pi {r^3} \cr & h = 2r \cr} $$